PAT (Advanced Level) Practice — 1132 Cut Integer (20 分)
题目链接:https://pintia.cn/problem-sets/994805342720868352/problems/994805347145859072
Cutting an integer means to cut a K digits lone integer Z into two integers of (K/2) digits long integers A and B. For example, after cutting Z = 167334, we have A = 167 and B = 334. It is interesting to see that Z can be devided by the product of A and B, as 167334 / (167 × 334) = 3. Given an integer Z, you are supposed to test if it is such an integer.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤ 20). Then N lines follow, each gives an integer Z (10 ≤ Z <231). It is guaranteed that the number of digits of Z is an even number.
Output Specification:
For each case, print a single line Yes
if it is such a number, or No
if not.
Sample Input:
3
167334
2333
12345678
Sample Output:
Yes
No
No
!注意:如果不判断两个分离数相乘的结果是否为0,则会出现浮点错误。
#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
int main(){
int n;
char a[11];
cin>>n;
for(int i=0;i<n;i++){
cin>>a;
int len=strlen(a);
ll m=0,m1=0,m2=0,m3=0;
for(int i=0;i<len;i++){
m=m*10+(a[i]-'0');
}
for(int i=0;i<len/2;i++){
m1=m1*10+(a[i]-'0');
}
for(int i=len/2;i<len;i++){
m2=m2*10+(a[i]-'0');
}
m3=m1*m2;
if(m3!=0&&m%m3==0){
printf("Yes\n");
} else{
printf("No\n");
}
}
return 0;
}
利用stoi :字符串转数值
#include<iostream>
#include<string>
using namespace std;
typedef long long ll;
int main(){
int n;
cin>>n;
string s;
for(int i=0;i<n;i++){
cin>>s;
int len=s.length();
ll m1,m2,m3,m;
m=stoi(s);
m1=stoi(s.substr(0,len/2));
m2=stoi(s.substr(len/2,len));
m3=m1*m2;
if(m3!=0&&m%m3==0){
printf("Yes\n");
}else{
printf("No\n");
}
}
return 0;
}